I already did that but I exactly have the same problem! :(
How it can be possible?
Thanks anyway!
[EMAIL PROTECTED] wrote:
What are you running this in ? In a standalone
client or in an application server (which one?)
Why don't you try using wsdl2java and create
stub classes to use instead. It is far simpler.
then your call to the web service just becomes:
ret =
echoWebService.test(code,utilizador_portal,password);
Hi there!
I receive this exception:
java.lang.ClassCastException:
org.apache.axis.transport.http.HTTPSender
,when I execute this code:
try {
String endpoint = "http://localhost:8080/axis/Echo.jws";
logger.info("DEBUG
>>> 1");
Service service =
new Service();
logger.info("DEBUG
>>> 2");
Call call = (Call)
service.createCall();
logger.info("DEBUG
>>> 3");
call.setTargetEndpointAddress( new java.net.URL(endpoint) );
call.setOperation("test");
call.addParameter("code", XMLType.XSD_STRING, ParameterMode.IN);
call.addParameter("utilizador_portal", XMLType.XSD_STRING,
ParameterMode.IN);
call.addParameter("password", XMLType.XSD_STRING, ParameterMode.IN);
call.setReturnType(XMLType.XSD_STRING);
logger.info("DEBUG
>>> 4");
//call.setOperationName(new QName("http://soapinterop.org/", "test"));
//logger.info("DEBUG >>> 5");
String ret =
(String) call.invoke(new Object[] {"123", "user", "passwd"});
<========= EXCEPTION!!
logger.info("DEBUG
>>> 6");
System.out.println("RESULT >>> '" + ret + "'");
logger.info("DEBUG
>>> 7");
} catch (Exception e) {
System.err.println(e.toString());
}
Any help? :(
---
Bruno Vg
|
- Re: exception when call.invoke Bruno Gonçalves
-
